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Relativistic particle decay

  • Thread starter Kruger
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  • #1
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Homework Statement



A particle with momentum p0, mass m0 and energy E0 decays into two particles with mass m1 and m2. Find the energy of the particle E1 and E2.

Homework Equations



Four-momentum!

The Attempt at a Solution



I calculated the energy of particle 1 in S' (system where particle 0 is at rest) in dependence of m0', m1', m2'. I got also the momentum of the particle 1 in system S' which I can write in dependence of m0', m1', m2' and E1'. For example I have taken the coordinates of system S' in such a way, that I can write the momentum p'1 of particle 1 as: p'1=(0,py',0) with py'=sqrt(f(m0',...)). Thus getting the four-momentum of particle 1: (E', 0, c*py', 0)

Well, what my question concerns. How can I translate the whole thing back into system S? Lorentz-Transformation or is there a easyer way? How can I go on to fullfill the task?
 

Answers and Replies

  • #2
Dick
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The problem can't be solved as stated. E1 and E2 aren't fixed by knowledge of the initial energy and masses alone.
 
  • #3
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Well, the diffraction angle of p1 relative to p0, the energy E0 and momentum p0 are given as well as the masses m0,m1,m2.
 
  • #4
Dick
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So you have an angle between p0 and p1, right? That's different... Then write the four momentum equation (E0,p0)-(E1,p1)=(E2,p2) and square both sides (in the Lorentzian sense). What do you get?
 
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  • #5
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Ok, this way I can solve it easy. It is no problem. I want to try it the other way. I mean, I know the solution of the problem in the frame of the moving particle p0 and want to transform it into the lab system (my system).
 
  • #6
Dick
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Ok, this way I can solve it easy. It is no problem. I want to try it the other way. I mean, I know the solution of the problem in the frame of the moving particle p0 and want to transform it into the lab system (my system).
Well then, take a specific solution in the rest frame, like p1 and p2 along the y-axis) and figure out what direction you have to boost in to get the required angle. Sounds masochistic to me though.
 
  • #7
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I have a question for solving a similar problem: A particle of mass M at rest decays into two particles of masses m1 and m2. Show that the totatl energy of the first particle is E1 = c(sqr) (M (sqr)+ m1 (sqr) - m(sqr) ) / (2M). and that E2 is obtained by interchanging m1 and m2.
Can you please give me a hint how can I start working it. I know I have to use the energy-momentum 4-vector but do I use only one component for P?
 
  • #8
Dick
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Both outgoing particles have the equal and opposite momenta - so you can assume there is only one component if you like. You know E1+E2=M and E^2-p^2=m^2 for each particle. So that's three equations. Eliminate p and E2 to get your expression. You'll have to fill in the c's - I set them to 1.
 
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  • #9
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I am still confused with this problem, I have the following equations:
E1+E2=MC^2 .................................(1)
E1^2-(p/C)^2=m1^2(C^4)...............(2)
E2^2+(p/C)^2=m2^2(C^4).............. (3) [p is equal in magnitude but oppsite in direction]

adding 2 and 3:

E1^2 + E2^2 = ( m1^2 + m2^2) C^4

Plugging the value of E2 from 1 gives a quadratic eguation:

2 E1^2 - 2MC^2 E1 + M^2C^4 = (m1^2 +m2^2 )C^4

Solving this equation doesn't give me the needed answer! Am I doing something wrong?
 
  • #10
Dick
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(-p/c)^2=(p/c)^2. The square of a negative is positive. You have a flipped sign.
 
  • #11
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Oh, I got it now! Thank you so so much!
 
  • #12
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I have the following problem: In a collision process a particle of mass m2, at rest in the lab frame is struck by a particle of mass m1 with energy E(L), and momentum P(L). In the centre of mass frame the two particles will have 3-momenta p, and -p respectively. Show that the Lorentz boost parametre B(CM) and describing the velocity of the centre of mass frame in the laboratory is: B(CM) = cP(L)/ (m2c^2+E(L))

I started my solution: B= V(CM)/C
P= [gamma](CM)*m2* V(CM)
E2= [gamma](cm)*m2*c^2
E1= [gamma](cm)*m1*c^2....... (not so useful)

now when trying to solve the first the two equations I get:

V(CM)= P*C^2/E2 , this means that E2 = m2c^2+ E(L)! Does this answer make sense? The energy of the second particle in CM frame is equal to total "spatial" energy in Lab frame! How can this be true?
 
  • #13
Dick
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I don't think you are doing a full lorentz boost, it should look more like this:

[tex]\left(\begin{array}{ccc}\gamma&-\beta \gamma\\-\beta \gamma&\gamma\end{array}\right)\left(\begin{array}{c}E\\p\end{array}\right) = \left(\begin{array}{c}E'&p'\end{array}\right)[/tex]

Apply this to (m2 0) and (EL PL) and set the resulting momenta to be negatives of each other and it's dead simple.
 

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